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7 EXAMPLE 1 Determine whether the series converges or diverges. 5n2 + 5n + 3 n = 1 SOLUTION For large n the dominant term in the denominator is 5n², so we compare the given series with the series E 7/(5n²). Observe that 7 7 5n2 + 5n + 3 5n because the left side has a bigger denominator. (In the notation of the Comparison Test, a, is the left side and b. is the right side.) We know that Σ |x Σ 7 5n2 n = 1 n2 n = 1 is convergent because it's a constant times a p-series with p = x > 1. Therefore Σ 5n2 + 5n + 3 n = 1 is convergent by the Comparison Test.

7 EXAMPLE 1 Determine whether the series converges or diverges. 5n2 + 5n + 3 n = 1 SOLUTION For large n the dominant term in the denominator is 5n², so we compare the given series with the series E 7/(5n²). Observe that 7 7 5n2 + 5n + 3 5n because the left side has a bigger denominator. (In the notation of the Comparison Test, a, is the left side and b. is the right side.) We know that Σ |x Σ 7 5n2 n = 1 n2 n = 1 is convergent because it's a constant times a p-series with p = x > 1. Therefore Σ 5n2 + 5n + 3 n = 1 is convergent by the Comparison Test.

College Algebra
College Algebra
7th Edition
ISBN:9781305115545
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter8: Sequences And Series
Section: Chapter Questions
Problem 6CC
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7 Determine whether the series S 5n2 EXAMPLE 1 converges or diverges. + 5n + 3 n = 1 SOLUTION For large n the dominant term in the denominator is 5n2, so we compare the given series with the series E 7/(5n2). Observe that 7 7 5n2 + 5n + 3 5n2 because the left side has a bigger denominator. (In the notation of the Comparison Test, a, is the left side and b, is the right side.) We know that Σ 7 1 |× £ = 5n2 n = 1 n2 n = 1 is convergent because it's a constant times a p-series with p = x > 1. Therefore Σ 7 5n2 + 5n + 3 n = 1 is convergent v by the Comparison Test.
Transcribed Image Text:7 Determine whether the series S 5n2 EXAMPLE 1 converges or diverges. + 5n + 3 n = 1 SOLUTION For large n the dominant term in the denominator is 5n2, so we compare the given series with the series E 7/(5n2). Observe that 7 7 5n2 + 5n + 3 5n2 because the left side has a bigger denominator. (In the notation of the Comparison Test, a, is the left side and b, is the right side.) We know that Σ 7 1 |× £ = 5n2 n = 1 n2 n = 1 is convergent because it's a constant times a p-series with p = x > 1. Therefore Σ 7 5n2 + 5n + 3 n = 1 is convergent v by the Comparison Test.
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